Talk:Kerr metric

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1 The Boyer/Lindquist coordinate chart metric is wrong
2 To Merge or not to Merge?
3 The spin parameter
4 Will the real monopole please stand up?
5 trying to use Kerr solution for rotating stars is doomed

[edit] The Boyer/Lindquist coordinate chart metric is wrong

I am certain that the metric as given in the section "Boyer/Lindquist coordinate chart" is wrong. I think that the $r 2 + a 2$ [in $(r 2 + a 2)sin 2 θ d φ 2$ ] needing to be squared and divided by ρ² is the only issue, but I can't be 100% sure without a chance to work it all through.

I admit that I could be wrong, but I am certain enough that there is a problem here that I would rather have the red flag thrown back in my face than let this go unchallenged. --EMS | Talk 17:08, 27 July 2005 (UTC)

Why are you certain? I didn't write this part of the article, but whoever did just copied (5.29) from Hawking & Ellis or (21.1) in Stephani or (8.32) in Ohanian & Ruffini, or the same expression as given in some other standard textbook. Even better, long years ago, I verified that this does give a vacuum solution.

I guess the confusion might have arisen if you have seen the terms in the BL line element collected in a different way (I'll give an example below), but all expressions for the BL line element should agree if you multiply everything out. Another possibility is that you saw the line element in a chart which looks like BL but isn't (for example, de Felice and Clarke introduce a "rotating" BL chart before they get to the standard BL chart).

This "controversy" may be moot since I've been planning to rewrite the article anyway to discuss the Kerr vacuum in much greater detail using various coframes. For example, from the line element as given in (19.27) in D'Inverno or (217) in Chandrasekhar we can read off the following coframe:

$\sigma^0 = -\frac{\delta}{\rho} \, \left( dt + a \sin(\theta)^2 d\phi \right)$

$\sigma^1 = \frac{\rho}{\delta} \, dr$

$\sigma^2 = \rho \, d\theta$

$\sigma^3 = \frac{\sin(\theta)}{\rho} \, \left( -a \, dt + (r^2 + a^2) d\phi \right)$

where

$\delta^2 = r^2 - 2 m r + a^2, \; \; \rho^2 = r^2 + a^2 \cos(\theta)^2$

Then, the metric is

$-\sigma^0 \otimes \sigma^0 + \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 + \sigma^3 \otimes \sigma^3$

This is one of the simpler coframes, but several others are of great interest. For example, the LNR coframe is given in the textbook by de Felice & Clarke. The Doran coframe is a generalization of the LeMaitre coframe for Schwarzschild. Other charts in common use include Eddington charts, Doran chart (generalization of Painleve chart), Plebanksi chart, Weyl canonical chart, the original Kerr chart, etc., and the coframes can be written in all of these charts. So you might see many different ways of writing down the Kerr solution.---CH (talk) 22:49, 27 July 2005 (UTC)

I don't think that it was copied correctly. I come up with <math>g_{\phi\phi} = \left [ \left (r^2 + a^2 \right ) + 2mra^2/\rho \right] \sin^2 \theta</math> from the metric as given in the article. That is not correct. --EMS | Talk 23:46, 27 July 2005 (UTC)<\s>

Ah ha! Now I see what I was missing: The

s i n 2 θ

was squaring also. Now it works out OK. This just is not a form of Kerr that I am used to. The coframe version and the spelled-out versions are ones that I familiar with. --EMS | Talk 23:54, 27 July 2005 (UTC)

[edit] Sorry for the confusion

I was exhausted yesterday, and did not realize just how tired I was. I just cannot do any intellectual "heavy lifting" when I'm like that. I knew that $(r 2 + a 2) 2 / ρ 2$ was involved with $g φφ$ . I had lost track of the fact that it was involved with the coframe version and sorts itself out into a form such as is in the article.

So I was tired and irritated (due to other things) and was not able to respond to my own self-red-flag of "I know I could be wrong". --EMS | Talk 22:03, 28 July 2005 (UTC)

[edit] Roy P. Kerr

Hi, Taxman, don't worry, I am certain the initial is indeed "P."---CH (talk) 20:11, 5 August 2005 (UTC)

[edit] Proposed biographical subcategory

There are many professional biographies on leading figures which have appeared in the journals over the years. I am pretty sure I have seen one on Kerr, for example (can't recall where). Since I have my hands full with writing more technical articles, would someone who lives near a good physics research libraray be interested in systematically searching for such biographies, looking them up, and writing wikibiographies? The biographical articles should cite the published biography as a source. I see someone has written a brief biography of Roy P. Kerr; to avoid cluttering up a category (this one) which is already large (and I plan to enlarge it considerably), I have created the new subcategory "Contributors to general relativity".

[edit] To Merge or not to Merge?

I'm against merging rotating black hole with this article. They are parent subjects, not the same one. The content of these articles are enough different to support so. nihil 09:25, 25 November 2005 (UTC)

Nihil may have a point. The "no hair theorem" is or should be surprising, and Price's theorem is often summarized by saying that any inhomogeneties or anisotropies outside a black hole which can be radiated away, will be radiated away (which restores the Kerr exterior vacuum of a relaxed rotating black hole). So, perhaps the article Rotating black hole can discuss these facts, while referring to Kerr vacuum for the relaxed state.---CH [[User_talk:Hillman|(talk)]] 18:26, 25 November 2005 (UTC)

[edit] The spin parameter

The explanation is incomprehensible as the number a = J/M is not dimensionless. Somebody please clarify. Bo Jacoby 08:56, 26 January 2006 (UTC)

units where c=1. If it troubles you, then use a=J/Mc². -lethe ^talk 09:11, 26 January 2006 (UTC)

Thanks! But the SI-unit of a = J/Mc² is (m²×kg/s)/(kg×m²/s²) = s. Still not dimensionless. Bo Jacoby 10:43, 26 January 2006 (UTC)

Hmm... you're totally right. I do apologize for answering so hastily and incorrectly. It is not dimensionless. And in fact, I guess it shouldn't be dimensionless, it should have the same dimension as r and s. I don't know what the problem is. -lethe ^talk 10:54, 26 January 2006 (UTC)

So that stuff was added by an anonymous editor. I now think it's mistaken, but I won't remove it unless I know what it's supposed to say. I suspect that what is meant is that a can range from 0 to m, rather from 0 to 1, and the anonymous had used a reference with a different notation convention. But I'm not sure, so.... -lethe ^talk 11:12, 26 January 2006 (UTC)

First of all a indeed is not dimensionless at all, nor should it be. In geometrized units (where c=G=1), it is a length, just as M and r are. Secondly, the angular momentum J must be an area so that a=J/M can also be a length. To get these geometrized lengths, you need to multiply a mass by G/c² and an angular momentum by G/c³. --EMS | Talk 15:53, 26 January 2006 (UTC)

P.S. The anon mentioned above was most likely CH, who sometimes forgets to log in. if not, it was someone else who is quite familiar with the Kerr metric. --EMS | Talk 16:01, 26 January 2006 (UTC)

Oops, I mean thanks! -lethe ^talk 18:16, 26 January 2006 (UTC)

FYI - Turns out that I was wrong about the anon being CH. (I should have guessed once I realized what the error was.) My apologies to Chris. Let's just say that this talk page seems to be jinxed for me. :-( --EMS | Talk 04:55, 28 January 2006 (UTC)

I'm the anon. Sorry about the error, I usually use the Kerr metric with M normalized to be 1, and measure everything in R_G 's. In those units a ranges from 0 to 1. By writing the M's in explicitly in the metric, yes, the range changes from 0 to M. What's there now is what I indended, if not what I actually wrote. Sfuerst 22:53, 17 February 2006 (UTC)

The angular momentum J defines a length J/Mc, and the mass M defines another length, the half Schwarzschild radius GM/cc . The ratio Jc/GMM is dimensionless. Is that the spin parameter? Bo Jacoby 12:49, 29 January 2006 (UTC)

Here is an easy way to read off the geometrized units of parameters appearing in metric coefficients (or in one-forms defining a cobasis field). Since metric coefficients are dimensionless in geometric units, a quantity such as m/r must be dimensionless. Since r has units of length, so must m. Angles and trig functions of angles should be dimensionless (as should the argument of any exponential). Thus, $a \, \sin(\theta)^2/\sqrt(r^2+a^2)$ must be dimensionless, so since $\sqrt(r^2+a^2)$ has units of length, so must a. Last but not least, Bo please note that angular momentum has geometrized units of area, just as you would expect from Newtonian theory! See geometrized units. ---CH 23:26, 17 February 2006 (UTC)

[edit] Will the real monopole please stand up?

I have elaborated slightly on this admittedly confusing issue. As I have noted elsewhere, WP requires articles on relativistic moments, moments in weak field gtr (including Weyl moments and the multi-index mass/momomentum/stress moments), and an improved article on multipole moments in Newtonian gravitation. Ultimately, the new section should be linked to this forthcoming articles and can perhaps then be shortened.---CH 22:24, 2 June 2006 (UTC)

[edit] trying to use Kerr solution for rotating stars is doomed

On the article page it says "Open problems

The Kerr vacuum is often used as a model of a black hole, but if we hold the solution to be valid only outside some compact region (subject to certain restrictions), in principle we should be able to use it as an exterior solution to model the gravitational field around a rotating massive object other than a black hole, such as a neutron star--- or the Earth. This works out very nicely for the non-rotating case, where we can match the Schwarschild vacuum exterior to a Schwarzschild fluid interior, and indeed to more general static spherically symmetric perfect fluid solutions. However, the problem of finding a rotating perfect-fluid interior which can be matched to a Kerr exterior, or indeed to any asymptotically flat vacuum exterior solution, has proven very difficult. "

No wonder it is difficult. The Kerr solution has only 2 independent parameters: mass and angular momentum. These determine all the gravitational multipole moments. But a rotating star has a complicated structure and its multipole moments will be controlled by the rotational (angular) velocity within it (which can vary with distance from the center and with latitude or angle from the plane of symmetry), the pressure and density variations with depth and possibly more things like internal magnetic fields. So the external field or various example rotating stars is more than a 2-parameter family. Whoever wrote that stuff should probably fix it. Carrionluggage 06:20, 17 October 2006 (UTC)

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